sm9 85

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or
instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States
Copyright Act without the permission of the copyright owner is unlawful.
_______________________________________________________________________
PROBLEM (*9.85) As shown in Fig. P9.85. a solid shaft AB is to transmit 20 kW at 180 rpm
from the motor and gear to pulley D, where 8 kW is taken off, and to pulley C where the remaining 12
kW is taken off. If all
τ
= 45 MPa, determine the required diameter d of the shaft AB according to the
maximum shear stress theory of failure.
Assumption: The ratios of the pulley tensions are to be 'T1 / T2 = 3 and T3 / T 4 = 3.
*SOLUTION
159 180 60 3
P
TfHz
f
===
GEAR E
Torque:
159(20) 1060
3
E
TNm==⋅
The tooth force:
1060
( ) 10.6
0.1
E
Ez E
T
FkN
R
=− =− =−
PULLY D
Torque:
159(8) 424
3
D
TNm==⋅
22 2
(3 )(0.15) 0.3TT T=− =
or
2112
1.41 4.23 ; 5.64TkNTkNTTkN==+=
Thus,
12 12
( ) 5.64sin30 2.82 ; ( ) 5.64cos30 4.88
oo
zy
TT kN TT kN+= = += =
PULLY C
Torque:
159(12) 636
3
C
TNm==⋅
44 4
(3 )(0.2) 0.4TT T=− =
or
4334
1.59 4.77 ; 6.36T kN T kN T T kN==+=
The load, moment, and torque diagrams are shown below.
From statics:
7.74 3.5
yy
AkNBkN==
0.36 7.42
zz
AkNBkN==
Continued on next slide

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or
instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States
Copyright Act without the permission of the copyright owner is unlawful.
Point C:
222 222
1.548 0.072 0.636 1.675
yz
M
MT kNm++= + + = ⋅
Point D:
222
2.1 0.216 1.06 2.362 kN m++= ⋅
Point E:
222
0.7 1.488 1.06 1.956 kN m++= ⋅
The critical point is thus just to the right of D. Apply Eq. (9.30):
13
3
6
16(2362)
[ ] 64.4 10 64.4
(45 10 )
dmmm
π
−
==×=
×
Use a
65 mm− diameter shaft.
x
y
A
C
B
M
y, kN⋅m x
x
M
z , kN⋅m
z
D
0.4 m
Bz
A
z
T, kN⋅m
x
Figure S9.85
636 N⋅m
(d)
(
a
)
4.88 kN
E
10.6 kN
1060 N
⋅
m
A
y
-0.216
6.36 kN
0.2 m
-1.488
By
424 N
⋅
m
2.1 0.7
(c)
1.06
1.548
(
b
)
-0.072
0.636